Question

a) Refrigerators and freezers remove heat from food by evaporating volatile liquids such as CF₄ and then recompressing the vapor back into a liquid. Calculate what mass of water at 0.0 deg C can be converted into ice at 0.0 deg C when 1.00 kg of CF₄ evaporates. (delta H_vap = 136 kJ/kg for CF₄₎

b) Water in a leather flask or a porous clay pot can be cooled in hot climates by evaporating water from the surface of the container. Calculate what mass of water can be cooled from 35 deg C to 20 deg C via the evaporation of 60 g of water

Answer 1

Ans. #A. Amount of heat absorbed during vaporization of 1 kg CF₄ = dHvap x Mass of CF₄

                                                = (136 kJ/ kg) x 1.0 kg

                                                = 136 kJ

# The heat is absorbed from liquid water by CF₄.

So, total heat lost from water = -136 kJ. The –ve sign indicates that heat is lost from water.

# Now, heat lost by water is given by-

q = m C                                 - equation 1

Where,

q = heat gained

m = mass

C = enthalpy of fusion of ice

Or,

            136 kJ = m x (333.55 kJ/ kg)

            Or, m = 136 kJ / (333.55 kJ/ mol)

            Hence, m = 0.407734 kg

Hence, required mass of water converted into ice at 0.0⁰C = 0.407734 kg

                                                = 407.734 g

#B. Total heat absorbed during valorization of 60.0 g water = dHvap x mass of water

                                    = (2257.0 J / g) x 60.0 g

                                    = 135420 J

# The heat absorbed during vaporization is extracted from liquid water to cool it.

So,

Amount of heat lost by water during cooling = 135420 J

Now,

Heat lost by water given by-

q = m s dT                            - equation 1

Where,

q = heat lost

m = mass

s = specific heat

dT = Final temperature – Initial temperature

Or,

            135420 J = m x (4.184 J g^-10C^-1) x (35.0 – 20.0)⁰C

            Or, 135420 J = m x 62.76 J g^-1

                Or, m = 135420 J / (62.76 J g^-1)

            Hence, m = 2157.74 g

Hence, required mass of water = 2157.74 g = 2.158 kg